Let PQR be a triangle with R(-1,4,2).

Question

Let $PQR$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the mid point of $PQ$. The distance of the centroid of $	rianglePQR$ from the point of intersection of the lines $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$ is

Accepted Answer

Since $M$ is the midpoint of $PQ$, $P+Q=2M$, so the centroid of $\triangle PQR$ is
$$\frac{P+Q+R}{3}=\frac{2M+R}{3}=\frac{(4,2,4)+(-1,4,2)}{3}=(1,2,2).$$

Intersection of the two lines: Line 1 has $x=2$ (fixed), $y=2t,\ z=-3-t$. Line 2: $x=1+s,\ y=-3-3s,\ z=-1+s$. Setting $x=2$: $s=1$, giving $(2,-6,0)$, which also satisfies line 1 ($t=-3$).

$$\text{Distance}=\sqrt{(1-2)^2+(2+6)^2+(2-0)^2}=\sqrt{1+64+4}=\sqrt{69}.$$

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